Problem on Molar solution
Can someone please help me in getting through this problem. 2.0 molar solution is acquired, when 0.5 mole solute is dissolved in: (i) 250 ml solvent (ii) 250 g solvent (iii) 250 ml solution (iv) 1000 ml solvent
Select the right answer of the question. The vapour pressure of two liquids P and Q are 80 and 600 torr, respectively. The total vapour pressure of solution obtained by mixing 3 mole of P and 2 mole of Q would be: (a) 140 torr (b) 20 torr (c) 68 torr (d) 72 torr
During processing of food, several chemicals are added to it to augment its shelf life and to make it more attractive as well. Main types of food addi
Choose the right answer from following. NaClO solution reacts with H2SO3 as,. NaClO + H2SO3→NaCl+ H2SO4. A solution of NaClO utilized in the above reaction contained 15g of NaClO per litre. The
When 5.85 g of NaCl (having molecular weight 58.5) is dissolved in water and the solution is prepared to 0.5 litres, the molarity of the solution is: (i) 0.2 (ii) 0.4 (iii) 1.0 (iv) 0.1
Explain the reason behind that the chemists have not created a periodic table of compounds?
The accuracy of your written English will be taken into account in marking. 1. (a) Identify the spectator ions in the following equation &nb
The boiling point of a solution of 0.11 gm of a substance in 15 gm of ether was found to be 0.1oC higher than that of the pure ether. The molecular weight of the substance will be (Kb = 2.16) (a) 148 &nbs
Superphosphate has the formulate: CaH4 (PO4)2 H2O calculate the percentage of Phosphorus in this chemical. Show your calculations
1) Chromium(III) hydroxide is highly insoluble in distilled water but dissolves readily in either acidic or basic solution. Briefly explain why the compound can dissolve in acidic or in basic but not in neutral solution. Write appropriate equations to support your answer. 2) Explain how dissolving t
20 g of hydrogen is present in 5 litre of vessel. Determine he molar concentration of hydrogen: (a) 4 (b) 1 (c) 3 (d) 2 Choose the right answer from above.
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